# Average number of photons

1. May 6, 2017

### Silviu

1. The problem statement, all variables and given/known data
Calculate the mean number of photons in a cavity at temperature T and the mean energy per photon.

2. Relevant equations
In the large volume limit, the log of grand canonical partition function is: $log(Z_g) = \frac{gV}{h^3}\int log(1-e^{(E-\mu)/kT})dp^3$, with g - spin degeneracy, E energy, p - momentum and $\mu$ - chemical potential.

Also the average number of particles is given by $N = \lambda \frac{\partial log(Z_g)}{\partial \lambda}|_{T,V}$, with $\lambda = e^{\mu/kT}$

3. The attempt at a solution

I calculated N using the above formula and I obtained: $N = \frac{gV}{h^3}\int \frac{1}{e^{(E-\mu)/kT}-1}dp^3$. This is for a normal Bose-Einstein gas. Now for a photon I take $\mu = 0$ and $E=h\nu$ and I plug in in this equation. Is this correct? If not, how should I proceed?

2. May 6, 2017

When you use the Bose function for the mean occupancy off a state is $\frac{1}{e^{\frac{hc}{\lambda k T}}-1}$ the chemical potential $\mu$ is set to zero for reasons that I'm not entirely sure. $\\$ Anyway, the photon mode counting can proceed in a couple of ways, but I like to use a method that F. Reif uses: $e^{ik_xx}=e^{ik_x (x+L_x)}$. This means $k_x L_x =n_x (2 \pi)$ and similarly for y and z. This means the number of photon modes is $\Delta N=\Delta^3 n=V \frac{\Delta^3 k}{(2 \pi)^3}$. Then $\Delta^3 k=4 \pi k^2 \, dk$ and $k=\frac{2 \pi}{\lambda}$. There is also a factor of 2 for photon polarization. You can write the expression entirely in terms of a spectral density $f(\lambda) d \lambda$. (We counted photon modes. Don't forget to include the Bose factor for mean occupancy of the state(mode).) $\\$ You then integrate over $d \lambda$ from $0$ to $+\infty$ to get the mean number of photons in the volume $V$ at temperature $T$. $\\$ Editing: Question is, is your integral identical to this, with polarization factor $g=2$, and I believe the answer is yes. You can use $E_p=\frac{hc}{\lambda}$ and work entirely with wavelength. A google of the subject shows the integral that you have can be evaluated in closed form, but it is a somewhat difficult one. The answer they give is $I=16 \pi \zeta(3) (\frac{kT}{hc})^3$, where $\zeta$ is the Riemann zeta function. $\zeta(3)=1.202$ (approximately).